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2 years ago

At what location does gravity play a role in moving tectonic plates

Physics

2 answers:

Ksju [112]2 years ago

8 0

At the edge of the plates.

Lena [83]2 years ago

7 0

Answer:

In subduction of one plate over another on the basis of weight

Explanation:

When two plates moving and colliding, the plate with higher mass will subduct below the plate with lower weigh and lighter would remain above. generally continental types of plates are lighter in weight and oceanic plates are heavier.

Ex:- The collision of Eurasian plate and Indian plate. Indian plate is below and Eurasian plate above and formation of Himalaya

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During a plane showcase, a pilot makes circular "looping" with a speed equal to the sound speed (340 m/s). However, the pilot ca

Dmitriy789 [7]

Answer:

1472.98 m

Explanation:

Data provided:

Speed of circular looping, v = 340 m/s

Acceleration, a = 8g

here,

g is the acceleration due to the gravity = 9.81 m/s²

Now,

the centripetal acceleration is given as,

$a=\frac{v^2}{r}$

r is the radius of the loop

on substituting the respective values, we get

$8\times9.81=\frac{340^2}{r}$

r = 1472.98 m

5 0

3 years ago

What is the maximum value of the magnetic field at a distance of 2.5 m from a light bulb that radiates 100 W of single-frequency

Anvisha [2.4K]

Answer:

$1.04\times 10^{-7}$ T

Explanation:

$IP$ = Power of the bulb = 100 W

$r$ = distance from the bulb = 2.5 m

$I$ = Intensity of light at the location

Intensity of the light at the location is given as

$I = \frac{P}{4\pi r^{2}}$

$I = \frac{100}{4(3.14) (2.5)^{2}}$

$I$ = 1.28 W/m²

$B_{o}$ = maximum magnetic field

Intensity is given as

$I = \frac{B_{o}^{2}c}{2\mu _{o}}$

$1.28 = \frac{B_{o}^{2}(3\times 10^{8})}{2(12.56\times 10^{-7})}$

$B_{o} = 1.04\times 10^{-7}$ T

7 0

3 years ago

Electrons and protons travel from the Sun to the Earth at a typical velocity of 3.83 ✕ 105 m/s in the positive x-direction. Thou

Leona [35]

Answer:

$F=2.84*10^{-26}N$ & -y direction

$F=2.84*10^{-26}N$ & +y direction

Explanation:

From the question we are told that:

Speed of electron $V_e=3.83 * 10^5 m/s$ +x direction

Earths magnetic field $B_e=3.04 * 10^-^8$ +z direction

Generally the equation for magnetic force $F_m$ is mathematically given by

$F=q(V_e*B_e)$

where

$q=1.6*10^{-19}c\\\=i*\=z=-\=j$

$F=1.6*10^{-19}(3.83 * 10^5 m/s*3.04 * 10^-^8)$

$F=-2.84*10^{-26}N \=j$

Magnitude & Direction

$F=2.84*10^{-26}N$ -y direction

Generally the equation for magnitude and direction of the magnetic force on an electron. is mathematically given by

$\=F'=-1.6*10^{-19}(3.83 * 10^5 m/s*3.04 * 10^-^8)$

$\=F'=-2.84*10^{-26}N \=j$

Magnitude & Direction

$F=2.84*10^{-26}N$ & +y direction

5 0

3 years ago

A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t

Yuliya22 [10]

Answer:

a) $W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

b) $W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

c) $V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

d) $d_1 =0.183m$ or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

$W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

Part b

For this case first we can convert the spring constant to N/m like this:

$2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}$

And the work donde by the spring on this case is given by:

$W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

$W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i$

And if we solve for the initial velocity we got:

$V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

$-1/2mV^2_i = W_g + W_{spring}$

And replacing we got:

$-1/2mV^2_i =mg d_1 -1/2 k d^2_1$

And we can put the terms like this:

$\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0$

If we multiply all the equation by 2 we got:

$k d^2_1 -2 mg d_1 -m V^2_i =0$

Now we can replace the values and we got:

$200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0$

$200 d^2_1 -2.058 d_1 -6.3525=0$

And solving the quadratic equation we got that the solution for $d_1 =0.183m$ or 18.3 cm because the negative solution not make sense.

5 0

2 years ago

Select all that apply.

GenaCL600 [577]

The correct answer to the question is: A) miles/hour and B) metre/ second.

EXPLANATION:

Before answering this question, first we have to understand speed.

The speed of a body is defined as the rate of distance travelled or the distance travelled by a body per unit time.

Hence, it is a derived quantity which is obtained from distance and time.

The unit of distance can be metre, miles, and the unit of time can be second, minutes or hour.

As speed is the distance covered per unit time, the perfect units will be miles/hour and metre/second.

Hence, the correct options are first and second.

5 0

3 years ago

Read 2 more answers